Math Problem Statement

The arm of a crane at a construction site is 19.0 m long, and it makes an angle of 13.2° with the horizontal. Assume that the maximum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane withstand if the maximum load the crane can handle is 706 N? Answer in units of N·m. What is the maximum load for this crane at an angle of 35.8° with the horizontal? Answer in units of N.

Solution

I see that you're working with a problem involving the crane's arm, torque, and load. The problem is split into two parts:

Part 1:

The arm of the crane is 19.0 m long and makes an angle of 13.2° with the horizontal. You are asked to find the maximum torque the crane can handle, given the maximum load the crane can carry is 706 N.

To calculate torque (τ), we use the formula:

τ=F×r×sin(θ)\tau = F \times r \times \sin(\theta)

where:

  • F=706NF = 706 \, \text{N} is the force (maximum load),
  • r=19.0mr = 19.0 \, \text{m} is the length of the crane's arm,
  • θ=13.2\theta = 13.2^\circ is the angle between the crane's arm and the horizontal.

Part 2:

The question asks for the maximum load when the angle changes to 35.8°. We can rearrange the torque equation to solve for the force:

F=τr×sin(θ)F = \frac{\tau}{r \times \sin(\theta)}

where:

  • τ\tau is the maximum torque from part 1,
  • rr and θ\theta will change accordingly.

Let me now compute both parts for you. math

Given values for part 1

F_max = 706 # Maximum load in Newtons r = 19.0 # Length of the crane arm in meters theta_1 = 13.2 # Angle in degrees for part 1

Converting angle to radians for calculation

theta_1_radians = math.radians(theta_1)

Calculating the torque for part 1

torque_max = F_max * r * math.sin(theta_1_radians)

Now for part 2, we need the new angle and will calculate the maximum load

theta_2 = 35.8 # New angle in degrees for part 2 theta_2_radians = math.radians(theta_2)

Solving for the maximum load in part 2

F_max_new = torque_max / (r * math.sin(theta_2_radians))

torque_max, F_max_newHere are the results for the two parts of the problem:

Part 1:

The maximum torque the crane can handle is approximately 3063.1 N·m.

Part 2:

The maximum load the crane can carry when the angle is 35.8° with the horizontal is approximately 275.6 N.

Let me know if you need further details or have any questions!

Follow-up Questions:

  1. How does torque depend on the angle between the crane arm and the horizontal?
  2. What would happen if the crane arm was shorter? How would that affect the maximum torque?
  3. What is the role of sine in the torque calculation?
  4. How can you modify the formula to calculate torque for a vertical crane arm?
  5. Can torque calculations be applied to other rotational systems like gears or pulleys?

Tip: To maximize torque, try increasing the length of the arm and the angle relative to the horizontal (but not too much, as the load might become unstable at high angles).

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Math Problem Analysis

Mathematical Concepts

Torque
Trigonometry
Force
Angles
Physics

Formulas

Torque (τ) = F × r × sin(θ)
Maximum Load (F) = τ / (r × sin(θ))

Theorems

Sine Law

Suitable Grade Level

Grades 11-12